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If it given that the eigenvectors corresponding to distinct eigenvalues of a matrix are orthogonal, and the eigenvalues are real, can I get the conclusion that the matrix is symmetric/self-adjoint?

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No, consider $$\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}.$$

If you add the condition that the eigenspaces sum to the entire space, the answer is...

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Then, would you please help me with this problem? It is given that a 2 by 2 matrix has eigenvalues of 0 and 1, and the corresponding eigenvectors are {{1},{2}} and {{2},{-1}}. How can I tell this is a symmetric matrix? Thanks. –  Scorpio19891119 Nov 2 '12 at 14:26
    
@Scorpio19891119: Did you read the last sentence in my answer? If the matrix has a full set of eigenvectors, which your matrix does, then the answer is yes. –  wj32 Nov 3 '12 at 5:08

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